Bijectivity of a trigonometric function

Question

Let $f : (-1,1)\to (-\pi/2,\pi/2)$ be the function defined by $f(x)= \tan^{-1}\left(\frac{2x}{1-x^2}\right)$ the verify that $f$ is bijective

To check objectivity I assumed 2 variables $x$ and $y$ to be equal and so as to prove $f(x)=f(y)$. But I couldn't do so. I also wish to prove surjectivity.

Answer 1

I guess that $f(x)= \arctan(\frac{2x}{1-x^2}).$ I am right ?

Now, if $x,y \in (-1,1)$ and $f(x)=f(y)$, then we have to show that $x=y.$ We get $\frac{2x}{1-x^2}=\frac{2y}{1-y^2}, $ since $\arctan$ is strictly increasing.

This gives $x-y=xy(y-x)$. Now suppose that $x \ne y.$ Then we have $xy=-1$. But this is impossible, since $x,y \in (-1,1)$, hence $x=y.$

Answer 2

Let $$ f(x)=\arctan\frac{2x}{1-x^2},\qquad g(x)=\frac{2x}{1-x^2} $$

The derivative is $$ f'(x)=\frac{1}{1+g(x)^2}g'(x) $$ Now $$ 1+g(x)^2=1+\frac{4x^2}{(1-x^2)^2}=\frac{(1+x^2)^2}{(1-x^2)^2} $$ and $$ g'(x)=2\frac{1-x^2+2x^2}{(1-x^2)^2} $$ which means that $$ f'(x)=\frac{(1-x^2)^2}{(1+x^2)^2}\frac{2(1+x^2)}{(1-x^2)^2}=\frac{2}{1+x^2} $$ which is positive. Since $$ \lim_{x\to-1^+}f(x)=-\frac{\pi}{2},\qquad\lim_{x\to-1^-}f(x)=\frac{\pi}{2} $$ you're done.

By the way, this shows that, for $x\ne\pm1$, $$ f(x)=\begin{cases} c_++2\arctan x & x>1 \\[4px] c_0+2\arctan x & -1<x<1 \\[4px] c_-+2\arctan x & x<-1 \end{cases} $$ Then $c_0=0$, by evaluating at $0$. Since the limits of $f$ at $\pm\infty$ are $0$, we get that $c_+=-\pi$ and $c_-=\pi$.

Answer 3

$$\forall x\in(-1,1):\left(\frac x{1-x^2}\right)'=\frac{1+x^2}{(1-x^2)^2}>1$$

and

$$\forall t:(\arctan(t))'=\frac1{1+t^2}>0.$$

Both functions are monotonous and continous.